A copper wire stretched so as to make it 0.1% longer. The percentage increase in the resistance of the wire is :

1.0

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2.0

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0.1

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0.2

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Solution

The correct option is C $0.2$
Resistance, $R = \frac{ρ l}{A} = \frac{ρ l^{2}}{V}$
Here volume $V$ remains same so $R \propto l^{2}$
Let initial length $l_{1} = l_{0}$ and final length $l_{2} = l_{0} + \frac{0.1}{100} l_{0} = 1.001 l_{0}$
thus, $\frac{R_{2}}{R_{1}} = \frac{l_{2}^{2}}{l_{1}^{2}} = \frac{(1.001 l_{0})^{2}}{l_{0}^{2}} = 1.002$
The % rise in resistance $= \frac{R_{2} - R_{1}}{R_{1}} \times 100 = [R_{2} / R_{1} - 1] 100 = [1.002 - 1] 100 = 0.2$ %

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